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Question Number 80650 by jagoll last updated on 05/Feb/20

$$\underset{x\to a}{\lim }\frac{{(\mid x\mid -a)}^3-{(\mid a\mid -a)}^3}{x-a}=P,a<0$$ $$\underset{x\to a}{\lim }\frac{{(\mid x\mid -a)}^2-{(\mid a\mid -a)}^2}{x^2-ax}=?$$

Commented byjohn santu last updated on 05/Feb/20

$$\underset{x\to a}{\lim }\frac{(\mid x\mid -a)-(\mid a\mid -a)}{x-a}=\frac{P}{3{(-2a)}^2}=\frac{P}{12a^2}$$ $$\Rightarrow \underset{x\to a}{\lim }\frac{(\mid x\mid -a)-(\mid a\mid -a)}{x-a}\times \underset{x\to a}{\lim }\frac{(\mid x\mid -a)+(\mid a\mid -a)}{x}=$$ $$\frac{P}{12a^2}\times \frac{-4a}{a}=-\frac{P}{3a^2}$$

Commented byjagoll last updated on 05/Feb/20

$$thxmisterjohn$$

Answered by MJS last updated on 05/Feb/20

$$P{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$

Commented byjagoll last updated on 05/Feb/20

$$why?$$

Commented byjagoll last updated on 05/Feb/20

$$i^{\prime }mcorrectmyquestionsir$$

Commented byMJS last updated on 05/Feb/20

$$\mathrm{for}\mathrm{the}\mathrm{corrected}\mathrm{question}$$ $$P=-12a^2$$ $$\mathrm{and}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{limit}\mathrm{is}4\mathrm{\forall }a<0\mathrm{which}\mathrm{is}-\frac{P}{3a^2}$$

Commented byjagoll last updated on 05/Feb/20

$$thankyoumister$$

Answered by MJS last updated on 05/Feb/20

$$a<0\Rightarrow x<0\mathrm{if}x\mathrm{close}\mathrm{enough}\mathrm{to}a$$ $$\left(1\right)\Rightarrow \frac{{(\mid x\mid -a)}^3-{(\mid a\mid -a)}^3}{x-a}=-x^2-4ax-7a^2\underset{x\to a}{=}$$ $$=-12a^2$$ $$$$ $$\left(2\right)\Rightarrow \frac{{(\mid x\mid -a)}^2-{(\mid a\mid -a)}^2}{x^2-ax}=\frac{3a}{x}+1\underset{x\to a}{=}4$$

Commented byjagoll last updated on 05/Feb/20

$$howtoget-x^2-4ax-7a^{2}mister?$$

Commented byMJS last updated on 05/Feb/20

$$r<0\Rightarrow \mid r\mid =-r$$ $$\frac{{(-x-a)}^3-{(-a-a)}^3}{x-a}=\frac{-{(x+a)}^3+{\left(2a\right)}^3}{x-a}=$$ $$=\frac{-x^3-3{ax}^2-3a^{2}x+7a^3}{x-a}=-\frac{(x-a)(x^2+4ax+7a^2)}{x-a}=$$ $$=-x^2-4ax-7a^2$$ $$$$

Commented byjagoll last updated on 05/Feb/20

$$ooiunderstandsir.thankmuch$$

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